Recent content by FSNxPhantom

Thank you for your help. I'll see what my professor did to get that answer.

Ex = E1 cos (49.46) = 3.38 x 10^4 E = 1.64 x 10^ 5 So: Angle = arccos (Ex/E) = 78.1 degrees

Angle is 90 + arccos(0.52/0.80) = 139.46 Then: a = (5.2 x 10^4) b = (2.0 x 10^5) c^2 = a^2 + b^2 - 2abcos(139.46) = 2.4 x 10^5 c^2 = a^2 + b^2 + 2abcos(139.46) = 1.64 x 10^5 (close enough lol) So how do I find the second part to the answer? [77.979 degrees]

E2______E l.../ l.../ l.../ l.../ l.../ l./ l/\ Angle = o1 + o2 = 90 P_\_______ l\ E1./ l..\./ Angle from Cos o = (.52/.8) (?) l...\ l...\ l...\ q1->..<-q2 And the angle between E1 and E2 is: o = Cos^-1(.52/.8) + 90 degrees (from E2 to line in middle) ? And for Ex and Ey (if there is) Ex = E1...

SOOOo... E2______E l.../ l.../ l.../ l.../ l.../ l./ l/\ Angle P_\_______ l\ E1./ l..\./ Angle from Cos o = (.52/.8) (?) l...\ l...\ l...\ q1->..<-q2 Where the magnitude of E1 = 5.2 x 10^4 N/C and E2 = 2.0 x 10^5 Am I doing it right so far?

P l\ l..\ l...\ l...\ l_____\ q1->...<-q2 Is this it? Because I missed the class lesson and have to figure it all out right now. And sorry for the bother since I'm clueless right now.

When i used that, I put in: c^2 = (5.2 x 10^4)^2 + (2.0 x 10^5) - (2 x (5.2 x 10^4) x (2.0 x 10^5) x (cos 49.46)) = 1.7 x 10^5 I found 49.46 by: Cos-1 (.52/.8) = 49.46 Still confused since it was wrong. Did i do it wrong? And if so, how do i found the angle for it being the answer: [1.613 x...

Homework Statement A charge q1 = +8.15 muC is at the origin, and a charge q2 = -3.70 muC is on the x-axis 0.52 m from the origin. Find the electric field strength at point P, which is on the y-axis 0.61 m from the origin. Answer = [1.613 x 10^5 N/C at 77.979 degrees] Homework Equations E =...